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If p is the length of the perpendicular ...

If p is the length of the perpendicular from the origin to the line `(x)/(a) + (y)/(b) = 1, "then prove that " (1)/(p^(2)) = (1)/(a^(2)) + (1)/(b^(2))`

Text Solution

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The given line is
`...x/a+y/b=1`
`impliesbx+ay=ab`
`impliesbx+ay-ab=0`
P is the length of perpendicular from origin to the line
`thereforeP=(|b(0)+a(o)-ab|)/(sqrt(d^(2)+a^(2)))`
`impliesP=(ab)/(sqrt(b^(2)+a^(2)))`
Squaring both sides,
`impliesP^(2)=(a^(2)b^(21))/(a^(2)+b^(2))`
`implies(1)/(P^(2))=(a^(2)+b^(2))/(a^(2)b^(2))`
`implies(1)/(P^(2))=(a^(2)+b^(2))/(a^(2)b^(2))`
`implies(1)/(P^(2))=(a^(2))/(a^(2)b^(2))+(b^(2))/(a^(2)b^(2))`
`implies(1)/(P^(2))=(1)/(a^(2))+(1)/(b^(2))`
Hence proved.
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